1.

If extent of dissociation of both KCl and BaCl_2 of some concentrations with identical value of alpha as 0.9, what is the ratio of their van't Hoff factors?

Answer»

Solution :van.t Hoff FACTOR is given as , `i=(1+(n-1))/1 ALPHA`
n=2 for KCL , i=1+(2-1) `alpha` = 1+ `alpha`
=1+0.95 = 1.95
n=3 for `BaCl_2` , i=1+(3-1) `alpha` = 1+2 `alpha`
=1+1.9=2.9
The ratio of van.t Hoff factor for KCl and `BaCl_2`
=1.95 : 2.9 =1:1.51


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